How to prove $\lim_{n\to \infty} \left(\frac{S_n}{n^{1-\alpha}}\right) = \frac{1}{1-\alpha}$?

Question

Let $-\infty<\alpha<1 $ and $S_n=\sum_{k=1}^n \left(\frac{1}{k^\alpha}\right)$

How to prove $\lim_{n\to \infty} \left(\frac{S_n}{n^{1-\alpha}}\right) = \frac{1}{1-\alpha}$

Would really appreciate your input :)